1.12616

a = \(\frac { -5 }{ 4 } \) and r = \(\frac { -1 }{ 4 } \)
S\(\infty\) = \(\frac { a }{ 1-r } =\frac { \frac { -5 }{ 4 } }{ 1-\left( \frac { -1 }{ 4 } \right) } =-1\)

22° 30'

Here cosec x = - 2 \(\Rightarrow\)sin x =-\(1\over2\),
which is negative, so x lies in third or fourth quadrant.
\(\therefore\) sin x =\(-{1\over2}=-sin 30^o\)
= sin (180° + 30°) or sin (360° - 30°)
= sin 210° or sin 330°
= \(sin {7x\over 6}or sin {11\pi\over 6}\)
Hence the principal solutions are  \({7\pi\over 6},{11\pi\over 6}\)
Now sin x=-sin\({\pi\over 6}\) 
\(\Rightarrow \) x= n\(\pi\)+(-1)ⁿ\(({7\pi\over6})\) where n\(\in\)Z.

(i) Firm B
(ii) Firm B.

\(\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! } \)
\(\Rightarrow \ \frac { 1 }{ 6! } +\frac { 1 }{ 7\times 6! } =\frac { x }{ 8\times 7\times 6! } \)
\(\Rightarrow \ \frac { 1 }{ 6! } [1+\frac { 1 }{ 7 } ] \ =\frac { 1 }{ 6! } [\frac { x }{ 8\times 7 } ]\)
\(\Rightarrow \ \frac { 8 }{ 7 } =\frac { x }{ 8\times 7 } \ \Rightarrow \ x=64\)

\(\text{Here} P(E)=\frac { 1 }{ 4 } ,P(F)=\frac { 1 }{ 2 } \)
\(\text{and } P(E\cap F)=\frac { 1 }{ 8 } \)
(i) we know that
\(P(E\cup F)=P(E)+P(F)-P(E\cap F)=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } =\frac { 2+4-1 }{ 8 } =\frac { 5 }{ 8 }\)
\(\text{P(not E and not F)}=P(\overline { E } \cap \overline { F } )\)
\(=P(\overline { E\cup F) } =1-P(E\cup F)\)
\(=1-\frac { 5 }{ 8 } =\frac { 8-5 }{ 8 } =\frac { 3 }{ 8 } \)

Let P(A) = x then P(B) = 3x
A and B are mutually exclusive exhaustive events
\(\therefore \ P(A\cup B)=P(A)+(B)=1\)
\(\therefore \ x+3x=1\Rightarrow 4x=1\Rightarrow x=\frac { 1 }{ 4 } \)
\(Thus\ P(A)=\frac { 1 }{ 4 } \)

y²=20x

Let A = {x : x is a prime number}
A' = U - A
= {x: x ∈ N} - {x: xis a prime number}
= {x : x ∈ N, x is not a prime number}
or {x : x is positive composite number and x = 1}

3\(\sqrt { 3 } \) units

The equation of the circle is
(x - h)² + (y - k)²=r²
(x - 2)² + (y + 3)² = r².....(i)
The given lines are 3x - 2y -1=0 ....(ii)
4x + y - 27=0 ...(iii)
Solving (ii) and (iii) we have
\(\frac { x }{ 54+1 } =\frac { y }{ -4+81 } =\frac { 1 }{ 3+8 } \Rightarrow \frac { x }{ 55 } =\frac { y }{ 77 } =\frac { 1 }{ 11 } \)
∴  x=\(\frac { 1 }{ 11 } \) x 55=5 and y=\(\frac { 1 }{ 11 } \) x 77=7
Putting these values of x andy in (i)
(5 - 2)² + (7 + 3)² =r² ⇒ r² = 9 + 100 ⇒\(\sqrt { 109 } \)
Putting value of r in
(x - 2)² + (y + 3)² = (\(\sqrt { 109 } \))²
⇒ x² + 4 - 4x + y² + 9 + 6y=109 ⇒ ​​​​​​​ X² + y² - 4x + 6y - 96 =​​​​​​​0
Thus required equation of circle is​​​​​​​
x² + y² - 4x + 6y - 96 =​​​​​​​0.

{(1, 1), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3) (2, 4), (2, 5), (3, 1), (3, 3,) (3, 4), (3, 5)}

\({ n }_{ 1 }=25,\bar { { x }_{ 1 } } =18.2.{ \sigma }_{ 1 }=3.25\)
\({ n }_{ 2 }=15\sum _{ i=1 }^{ 15 }{ { x }_{ 1 } } =279\)
\(and\ \sum _{ i=1 }^{ 15 }{ { x }_{ 1 }^{ 2 } } =5524\)
\(for\ first\ set,\sum { { x }_{ 1 } } =25\times 18.2=455\)
\({ \sigma }_{ 1 }^{ 2 }=\cfrac { \sum { { x }_{ 1 }^{ 2 } } }{ 25 } -(18.2{ ) }^{ 2 }\)
\(\Rightarrow (3.25{ ) }^{ 2 }=\cfrac { \sum { { x }_{ 1 }^{ 2 } } }{ 25 } -331.24\)
\(\Rightarrow \sum { { { x }_{ 1 }^{ 2 }=25\times (10.5625+331.24) } }\)
\(=25\times 341.8025=8545.0625\)
\(For\quad combained\quad SD\quad of\quad the\quad 40\quad observations,N=40\)
\(Now,\sum { { x }_{ 1 }^{ 2 } } =5524+8545.0625=14069.0625\)
\(and\sum { { x }_{ 1 }=445+279=734 } \)
\(SD=\sqrt { \frac { 14069.0625 }{ 40 } -\left( \frac { 734 }{ 40 } \right) }\)
\(=\sqrt { 351.726-(18.35{ ) }^{ 2 } }\)
\(=\sqrt { 351.726-336.7225 } \)
\(=\sqrt { 15.0035 } =3.87\)

The given compound statement is S : 80 is a multiple of 5 and 4.
The component statements of the given compound statement are
p : 80 is a multiple of 5.
and q : 80 is a multiple of 4.
Then, \(S\equiv p\wedge q\)
We know that, 80 is mutiple of 5 as well as 4.
Therefore, the statements p and q are true.
Hence, the compound statement  \(S\equiv p\wedge q\)   is also true
i.e.  \(S\equiv p\wedge q\)  is a valid statement.

Mridul is neither cruel nor strict.

\(\lim _{h \rightarrow 0} \frac{1+h-1}{\log _{e}(1+h)}=\frac{1}{\lim _{h \rightarrow 0} \frac{\ln (1+h)}{k}} A n s .1\)

\(\frac { d }{ dx } (\frac { { x }^{ 5 }-cosx }{ sinx } )\)
\(=\frac { sinx\frac { d }{ dx } ({ x }^{ 5 }-cosx)-({ x }^{ 5 }-cosx)\frac { d }{ dx } sinx }{ \left( sinx \right) ^{ 2 } } \)
\(=\frac { sinx({ 5x }^{ 4 }+sinx)-({ x }^{ 5 }-cosx)cosx }{ { sin }^{ 2 }x } \)
\(=\frac { { 5x }^{ 4 }sinx-{ x }^{ 5 }cosx+1 }{ { sin }^{ 2 }x } \)

Let YZ-plane divides the line segment joining the points A (4, 8, 10) and B (6, 10, -8) at P (x, y, z) in the ratio k : 1.
Then, coordinates of P = \(\left( \frac { 4+6k }{ k+1 } ,\frac { 8+10k }{ k+1 } ,\frac { 10-8k }{ k+1 } \right) \) 
\(\left[ \therefore coordinates\quad of\quad internal\quad ration,\quad \left( \frac { { m }_{ 2 }{ x }_{ 1 }+{ m }_{ 1 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 2 }{ y }_{ 1 }+{ m }_{ 1 }{ y }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 2 }z_{ 1 }+{ m }_{ 1 }{ z }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \right] \) 
Since, P lies on the YZ-plane, so its x-coordinate is zero,
\(\therefore \)  \(\frac { 4+6k }{ k+1 } =0\Rightarrow k=-\frac { 2 }{ 3 } \) 
Therefore, YZ-plane divides AB externally in the ratio 2 : 3
Note: If k is negative, then given points are divided externally in the ratio k : 1.

Let the given point be A(2, 3), with lie on the line x-y+1=0

Let x - y + 1 = 0 intersect 2x - 3y + 9 = 0 at point B.
Then, required distance = length of AB = \(4\sqrt { 3 } units\)

x + \(\sqrt { 3 } \)y = 14

We know that,  AM > GM
\(\therefore \frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } >\sqrt { { a }^{ 2 }{ b }^{ 2 } } \Rightarrow \frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } >ab\)  ...(i)
\(similarly, \frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } >\sqrt { { b }^{ 2 }{ c }^{ 2 } } \Rightarrow \frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } >bc\)  ...(ii)
\(and \ \frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >\sqrt { c^{ 2 }{ a }^{ 2 } } \Rightarrow \frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >ca\)  ...(iii)
On adding Eqs. (i) and (iii), we get
\(\frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 }+{ c }^{ 2 } }{ 2 } +\frac { { c }^{ 2 }+{ a }^{ 2 } }{ 2 } >ab+bc+ca\)
\(\Rightarrow \)  a² + b² + c² > ab + bc + ca
Hence proved              

Let (r+1)th term involve y⁶.
Then, 
\(T_{ r+1 }=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad ({ 2y }^{ 2 })^{ 11-r }\quad \left( \frac { 3 }{ y } \right) ^{ r }\)
\(=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad 2^{ 11-r }\quad (y^{ 2 })^{ 11-r }.\quad 3^{ r }.y^{ -6 }\)
\(=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad 2^{ 11-r }\quad (3)^{ r }\times y^{ 22-3r }\)
Since, T_r+1 involve y⁶ , therefore 22-3r = 6
\(\Rightarrow 3r=16\Rightarrow r=\frac { 16 }{ 3 } ,\) which is not an integer.
Hence, there is no term containing y⁶.
Hence proved.

\(= { \frac { ^{ 15 }C_{ r } }{ ^{ 15 }C_{ r-1 } } }=\frac { 11 }{ 5 } \Rightarrow \frac { 15! }{ (15-r)!r! } \times \frac { (15-r+1)!\times (r-1)! }{ 15! } =\frac { 11 }{ 5 } \)
\(\Rightarrow \frac { (16-r) }{ r } =\frac { 11 }{ 5 } \Rightarrow 80-5r = 11r\)
16r = 80 ⇒ r = 5

We have,  \(|x-2|\ge 6\)
\(\Longrightarrow \)\(x-2\ge 6\)  or  \(x-2\le -6\) \(\left[ \because |x|\ge a\Longrightarrow x\ge a\quad or\quad x\le -a \right] \)
\(\Longrightarrow \)  \(x\ge 8\quad or\quad x\le -4\)
\(\Longrightarrow \) \(x\) \(\in \) [8, \(\infty \)) or \(x\)\(\in \) (-\(\infty \), 4]
\(\Longrightarrow \)  \(x\) \(\in \) [8, \(\infty \)) \(\cup \) (-\(\infty \), 4]
Hence, the required solution set is (-\(\infty \), 4] \(\cup \) [8, \(\infty \)).

\(\frac { 3\pi }{ 4 } \)

Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

Here, the given statement is true for all n \(\in\) N, therefore it is true for n=1 also
So, we have P(1) : 49+1++K is divisible by 64,
i.e. 65+k is divisible be 64
Clearly, k should be -1  [\(\therefore \)65-1=64 is divisible by 64]

We have \(f\left( x \right) =\frac { 1 }{ \sqrt { x+\left| x \right| } } \)
We know that,  \(\left| x \right| =\begin{cases} x,\quad ifx\ge 0 \\ -x,ifx<0\quad \end{cases}\)
\(\therefore \quad x+\left| x \right| =\begin{cases} x+x,\quad ifx\ge 0 \\ x-x,ifx<0\quad \end{cases}\)
\( \Rightarrow x+\left| x \right| =\begin{cases} 2x,\quad ifx\ge 0 \\ 0,ifx<0\quad \end{cases}\)
Here, the function \(f\left( x \right) =\frac { 1 }{ \sqrt { x+\left| x \right| } } \) assumes real values, if \(x+\left| x \right| >0\)
\(\Rightarrow 2x>0\ \Rightarrow x>0\quad [using\quad Eq.(i)]\)
\(x\in \left( 0,\infty \right) \)
Hence, domain of \(f=\left( 0,\infty \right) \)
Note:The domain of modulus function is always R i.e.\(\left( -\infty,\infty \right) \).

Let E and F denote the Indian nutrition food and fast food 
Then,  \(n(E\cup F)=65,n(E)=40\quad and\quad n(E\cap F)\quad =10\)
 \(\because \) \(n(E\cup F)=n(E)+n(F)-n(E\cap F)\)
\(\therefore \) 65 = 40 + n(F)-10 \(\Rightarrow \) n(F)=35
\(\therefore \) Only like fast food = \(n(E)-n(E\cap F)\)= 35  10 = 25 
They must balance the diet.

\(\frac { 12 }{ 13 } \)

Let us make the following table from the given data.

Hence,  \(N=\Sigma { f }_{ i }=7\) and \(\Sigma { f }_{ i }{ x }_{ i }=22A\)
\(\therefore \bar { x } =\frac { \sum _{ i=1 }^{ 6 }{ { f }_{ i }{ x }_{ i } } }{ N } =\frac { 22A }{ 7 } \)
Now, variance \(=\frac { 1 }{ N } \sum _{ i=1 }^{ 6 }{ { f }_{ i }({ x }_{ 1 }-\bar { x } )^{ 2 } } \)
\(\Rightarrow \ 160=\frac { 1 }{ 7 } \left( \frac { 1120 }{ 49 } \right) { A }^{ 2 }\)
[ \(\because \) variance = 60, given]
Ans. A=7 

\(f\prime (x)=\lim_ { h\rightarrow 0 }{ lim } \frac { { e }^{ 2x+2h }-{ e }^{ 2x } }{ h } =\lim_ { h\rightarrow 0 }{ lim } \frac { { e }^{ 2x }({ e }^{ 2h }-1) }{ 2h } x2\)
\(=2e^{ 2x }\)

\(\left( \frac { -3 }{ 10 } ,\frac { 5 }{ 2 } ,\frac { -1 }{ 5 } \right) \)

\({ (x-2) }^{ 2 }+{ (y-2) }^{ 2 }=\frac { { 2 }^{ 2 }{ (x+y-9) }^{ 2 } }{ { 1 }^{ 2 }{ +1 }^{ 2 } } \)
\(\Rightarrow \)  x² + y² - 4x - 4y + 8 = 2(x² + y² + 81 + 2xy -18x)
x² + 4xy + y² - 32x - 32y + 154 = 0

First, find out the values of AB,BC,CA by distance formula and prove that \(\triangle ABC\) is an isosceles triangle, then prove that this triangle is also right-angled triangle by Converse of Pythagoras theorem.

S_q=q³

General term in \({ \left( { x }^{ 2 }+\frac { 1 }{ x } \right) }^{ n }\)= ⁿC_r (x²)^n-r \({ \left( \frac { 1 }{ x } \right) }^{ r }\) 
for term independent of x, x^2n-2r-r = x⁰ \(\Rightarrow \) r = \(\frac { 2 }{ 3 } n\) 
Required term = ⁿC_r = \(\frac { n! }{ r!\left( n-r \right) ! } \)= \(\frac { n! }{ \left( \frac { n }{ 3 } \right) !\left( \frac { 2n }{ 3 } \right) ! } \)

Total number of points = m+n+k

This can give ^m+n+kC₃ number of triangles. But m points on l_1, taking 3 at a time, gives ^mC₃ combinations which produce no triangle.
Similarly, ⁿC₃ and ^kC₃  number of triangle cannot formed.
\(\therefore \) Required number of triangles        
= ^m+n+kC₃ - ^mC₃ - ⁿC₃ - ^kC₃

\(We\ have,\ 5{ x }^{ 2 }-4ix+9=0\)
\(a=5,\quad b=-4i\quad and\quad c=9\)
\(\therefore \ x=\frac { -(-4i)\pm \sqrt { { (-4i) }^{ 2 }-4\times 5\times 9 } }{ 2\times 5 } \)
\(=\frac { -4i\pm \sqrt { { (-4i) }^{ 2 }-4\times 5\times 9 } }{ 2\times 5 }\)
\(=\frac { 4i\pm \sqrt { -16-180 } }{ 10 } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { 4i\pm \sqrt { -196 } }{ 10 } ,\quad x=\frac { 4i\pm 14i }{ 10 } \)

We have, \(\sqrt { -25 } +3\sqrt { -4 } +2\sqrt { -9 } \)
\(=\sqrt { 25 } \sqrt { -1 } +3\sqrt { 4 } \sqrt { -1 } +2\sqrt { 9 } \sqrt { -1 }\)
= 5 x i + 3 x 2 x i + 2 x 3 i
= 5i + 6i + 6i
= 17i

\(Let\quad A=\frac { \pi }{ 7 } ,then\)
\(LHS=cos\frac { \pi }{ 7 } cos\frac { 2\pi }{ 7 } cos\frac { { 2 }^{ 2 }\pi }{ 7 } \frac { sin{ 2 }^{ 3 }A }{ { 2 }^{ 3 }sinA }\)
\(=\frac { sin8A }{ 8sinA } =\frac { sin(7A+A) }{ 8sinA } \)
\(=\frac { sin(\pi +A) }{ 8sinA } =\frac { -sinA }{ 8sinA } \quad [\because A=\frac { \pi }{ 7 } ]\)

g is a function; \(\alpha =2,\quad \beta =-1\)

(i) A triangle is equiangular implies that it is an obtuse angled triangle.
(ii) A triangle is equiangular only, if it is an obtuse angled triangle.
(iii) For a triangle to be equiangular, it is necessary that it is an obtuse angled triangle.
(iv) For a triangle to be obtuse angled triangle, it is sufficient that it is equiangular.
(v) If a triangle is not obtuse angled triangle, then it is not an equiangular triangle.

2k<(k+2)!\(\Rightarrow \)2k+2(k+2)!+2]
\(\Rightarrow \) (k+1)² < 2^k + 2^k    [(2k+1)<2^k for k\(\le \)3]
\(\Rightarrow \) (k+1)² < 2^k+1

Let \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ (n+1) } \)
For n =1
 P(1) = \(\frac { 1 }{ 1(1+1) } =\frac { 1 }{ 1+1 } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ (k+1) } \)   ...(i)
For n = k+1
∴ P(k+1)= \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } +\frac { 1 }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)
= \(\frac { k }{ k+1 } +\frac { 1 }{ (k+1)(k+2) } =\frac { k(k+2)+1 }{ (k+1)(k+2) } =\frac { { k }^{ 2 }+2k+1 }{ (k+1)(k+2) } \)       [Using (i)]
=  \(\frac { (k+1)^{ 2 } }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)= P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

The three examples of sentences, which are not statements, are as follows.
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

\(\overset{Lt}{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{\frac{1}{cotx}}\)=\(\overset{Lt}{x\rightarrow 0} [\frac{tan\frac{\pi}{4}+tanx}{1-tan\frac{\pi}{4}.tanx}]^{cotx}\)
=\(\overset{Lt}{x\rightarrow 0}[\frac{1+tanx}{1-tanx}]^{\frac{1}{tanx}}\) ⇒ \(\frac { \underset { \underset { tanx\rightarrow 0 }{ x\rightarrow 0 } }{ Lt } { \left[ [1+tanx \right] }^{ \frac { 1 }{ tanx } } }{ \underset { \underset { -tanx\rightarrow 0 }{ x\rightarrow 0 } }{ Lt } { \left[ { \left( 1-tanx \right) }^{ -\frac { 1 }{ taanx } } \right] }^{ -1 } } =\frac { e }{ { e }^{ -1 } } ={ e }^{ 2 }\)
Hence \(\overset{Lt}{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{cotx}=e^{2}.\)

Here n = 7, which is odd.
So the middle terms are \(\left( \frac { 7+1 }{ 2 } \right) th,\left( \frac { 7+1 }{ 2 } +1 \right) th\) are 4^th and 5^th terms.
The general term in the expansion of\(\left( 3-\frac { { x }^{ 3 } }{ 6 } \right) ^{ 7 }\) is
\({ T }_{ r+1 }=^{ 7 }{ C }_{ r }{ (3) }^{ 7-r }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ r }....(i)\)
Putting r = 3 and 4 in (i)
\(\therefore \quad { T }_{ 4 }=^{ 7 }C_{ 3 }{ (3) }^{ 7-3 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 3 }{ (3) }^{ 4 }.{ (-1) }^{ 3 }.\frac { { x }^{ 9 } }{ { (6) }^{ 3 } } \)
\(=35\times 81\times -\frac { { x }^{ 9 } }{ 216 } =-\frac { 105 }{ 8 } { x }^{ 9 }\)
Now \({ T }_{ 5 }=^{ 7 }C_{ 4 }{ (3) }^{ 7-4 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 4 }{ (3) }^{ 3 }.(-1)^{ 4 }\frac { { x }^{ 12 } }{ { (6) }^{ 4 } } \)
\(=35\times 27\times \frac { { x }^{ 12 } }{ 1296 } =\frac { 35 }{ 48 } { x }^{ 12 }\) 

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through point (2, 3).
\(\therefore(2-h)^{2}+3^{2}=25 \)
\(\Rightarrow(2-h)^{2}=25-9 \)
\(\Rightarrow(2-h)^{2}=16 \)
\(\Rightarrow 2-h=\pm \sqrt{16}=\pm 4 \)
\(\text { If } 2-h=4, \text { then } h=-2 . \)
\(\text { If } 2-h=-4, \text { then } h=6 .\)
Equation of required circle is
(x - 6)² + (y - 0)² = (5)²
⇒ x² + 36 - 12x + y² = 25
⇒ x² + y² - 12x + 11 = 0
When h=-2
Equation of required circle is
(x + 2)² + (y - 0)² = (5)²
⇒  x² + 4 + 4x + y² = 25
⇒  x² + y² + 4x - 21 = 0

x=2n\(\pi +{\pi\over 4} \ or \ x=2n \pi,n \in z\)

Let A(2, 0, 3), B(O, 3, 2) and C(O, 0,1) be given points.
Let P(x, y, 0) be any point in XY plane such that PA = PB = PC.
Now PA = PB => PA2 = PB2
\(\therefore \) (x - 2)² + (y - 0)² + (- 3)²
= (x - 0)² + (y - 3)² + (- 2)²
=> x²+4-4x+y²+9
=x²+y²+9-6y+4
=> 4x - 6y = 0 => 2x - 3y = 0 ....(i)
Also PB = PC => PB² = PC²
(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 2)²
\(\therefore \)(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 1)²
=> x²+ y² + 9 - 6y + 4
=x²+y²+1
=> 6y = 12 => y = 2
Putting value of y in (i), we have
2x-3x2=0 => x=3
Thus co-ordinates ofrequired point are (3, 2, 0).

From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.
Thus, required number of ways  = 
\({ }^{5} \mathrm{P}_{4}=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\)
= 1x 2 x 3 x 4 x 5 = 120
Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.
The number of ways in which units place is filled with digits is 2.
Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.
Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.
Number of ways of filling the remaining places \(={ }^{4} \mathrm{P}_{3}=\frac{4 !}{(4-3) !}=\frac{4 !}{1 !}\)
= 4 × 3 × 2 × 1 = 24
Thus, by multiplication principle, the required number of even numbers is = 24 × 2 = 48

\(\frac { 1+3i }{ 1-2i } \times \frac { 1+2i }{ 1+2i } =\frac { 1+2i+3i+6{ i }^{ 2 } }{ 1-4{ i }^{ 2 } } \)
=\(\frac { -5+5i }{ 5 } \)=-1+i
Let z = -1+i=r(cos\(\theta \)+i sin\(\theta \))
⇒ r cos\(\theta \)=-1 and r sin\(\theta \) =1   ...(i)
Squaring both sides of (i) and adding
r²(cos²\(\theta \)+sin²\(\theta \))=1+1
⇒  r²=2 ⇒ r= \(\sqrt { 2 } \)
∴ \(\sqrt { 2 } \)cos\(\theta \)=-1 and \(\sqrt { 2 } \)sin\(\theta \)=1
⇒ cos\(\theta \) =\(\frac { -1 }{ \sqrt { 2 } } \) and sin \(\theta \)=\(\frac { -1 }{ \sqrt { 2 } } \)
Since sin\(\theta \) is positive and cos\(\theta \) is negative
∴ \(\theta \) lies in second quadrant
∴    \(\theta \)=\(\pi -\frac { \pi }{ 4 } =\frac { 3\pi }{ 4 } \)
Hence polar form of z is \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } +isin\frac { 3\pi }{ 4 } \right) \)

\(\text { The equations of the given lines are }\)
\(2 x-3 y+4=0 \ldots(1) \)
\(3 x+4 y-5=0 \ldots(2) \)
\(6 x-7 y+8=0 \ldots \text { (3) }\)
\(\text { The person is standing at the junction of the paths represented by lines (1) and } (2)\)
\(\text { On solving equations }(1) \text { and }(2), \text { we obtain } x=-\frac{1}{17} \text { and } y=\frac{22}{17} \text { . }\)
\(\text { Thus, the person is standing at point }\left(-\frac{1}{17}, \frac{22}{17}\right) \text { . }\)
\(\text { The person can reach path (3) in the least time if he walks along the }\)
\(\text { perpendicular line to (3) from point }\left(-\frac{1}{17}, \frac{22}{17}\right) \text { . }\)
\(\text { Slope of the line }(3)=\frac{6}{7}\)
\(\text { Slope of the line perpendicular to line }(3)=-\frac{1}{\left(\frac{6}{7}\right)}=-\frac{7}{6}\)
\(\text { The equation of the line passing through and having a slope of }-\frac{1}{6} \text { is given } by\)
\(\left(y-\frac{22}{17}\right)=-\frac{7}{6}\left(x+\frac{1}{17}\right) \)
\(6(17 y-22)=-7(17 x+1) \)
\(102 y-132=-119 x-7 \)
\(119 x+102 y=125\)
\(\text { Hence, the path that the person should follow is } 119 x+102 y=125\)

The integers from 1 to 100 which are divisible by 2 are:
2, 4, 6, ..., 100
Here a = 2, d = 4 - 2 = 2 and a_n = 100
\( \because\) a_n = a + (n - l)d
\(\therefore\) 100 = 2 + (n - 1)²
\(\Rightarrow\) 100 - 2 = (n - 1)²
\(\Rightarrow\) (n-1) = \(\frac { 98 }{ 2 } =49\)
n = 49 + 1 = 50 
Now S₁ = \(\frac { 50 }{ 2 } =[2+100]\)
= 25 x 102 = 2550
Now, the integers from 1 to 100 which are divisible by 5 are 5, 10, 15, ..., 100.
Here, a = 5, d = 10 - 5 = 5 and a_n = 100
\(\because \)  a_n = a + (n - l)d
\(\therefore\)  100 = 5 + (n - 1) x 5
\(\Rightarrow\) 100 - 5 = (n - 1) x 5
\(\Rightarrow\) 95 = (n - 1) x 5
\(\Rightarrow\) (n-1) = \(\frac { 95 }{ 5 } =19\) \(\Rightarrow\) n = 20
\(\therefore\) S₂ = \(\frac { 20 }{ 2 } =[5+100]\)
= 10 x 105 = 1050
Also, the integers from 1 to 100 which are divisible by both 2 and 5 are:
10, 20, 30, ..., 100
Here, a = 10, d = 20 - 10 = 10 and a_n = 100
8ince, a_n = a + (n - l)d
 \(\therefore\)  100 = 10 + (n - 1)10
\(\Rightarrow\) (n-1) =\(\frac { 100-10 }{ 10 } =9\)
\(\Rightarrow\) n =9 + 1 = 10
Now, S₁₀ = \(\frac { 10 }{ 2 } \left[ 10+100 \right] \)
= 5x 100 = 550
Thus, the sum of the required integers from

Mean \(\left( \bar { x } \right) =A+\frac { \sum { fu } }{ N } \times h=27.5+\frac { 8 }{ 100 } \times 5=27.5+0.4=27.9\)
Standard deviation \(\left( \sigma \right) \) \(=\frac { h }{ N } \sqrt { N\sum { { fu }^{ 2 }-{ (\sum { fu) } }^{ 2 } } } \)
\(\sigma =\frac { 5 }{ 100 } \sqrt { 100\times 150-{ (8) }^{ 2 } } =\frac { 1 }{ 20 } \sqrt { 15000-64 } =\frac { 1 }{ 24 } \times 122.21=6.11\)
\(\therefore \ C.V.=\frac { \sigma }{ x } \times 100=\frac { 6.11 }{ 27.9 } \times 100=21.89\)

Here the sample space S = {I, 2, 3,4,5,6}
\(\therefore \) n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5} \(\Rightarrow \) n(A) = 3
\(Thus\ P(A)=\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event of getting a number greater than or equal to 3
B = {3, 4, 5, 6} \(\Rightarrow \) n(B) = 4
\(Thus\ P(B)=\frac { n(B) }{ n(S) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Let C be the event of getting a number less than or equal to 1
C = {I} \(\Rightarrow \) n(C) = 1
\(Thus\ P(C)=\frac { n(C) }{ n(S) } =\frac { 1 }{ 6 } \)
(iv) Let D be the event of getting a number more than 6
\(D=\phi \Rightarrow n(D)=0\)
\(Thus\ P(D)=\frac { n(D) }{ n(S) } \frac { 0 }{ 6 } =0\)
Let E be the event of getting a number less than 6
E = {I, 2, 3, 4, 5} \(\Rightarrow \) n(E) = 5
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

(i) \(\Rightarrow\) (ii)
A - B = {x : x ∊ A and x ∉ B}
Since A ⊂ B
\(\therefore\) A-B= ф
(ii) \(\Rightarrow\) (iii)
A-B= ф \(\Rightarrow\) A⊂B \(\Rightarrow\) A\(\cup\)B=B
(iii) \(\Rightarrow\) (iv)
A\(\cup\)B=B \(\Rightarrow\) A⊂B \(\Rightarrow\) A\(\cap\)B=A
(iv) \(\Rightarrow\) (i)
A\(\cap\)B=A\(\Rightarrow\) A⊂B
Thus (i) \(\Leftrightarrow\) (ii) \(\Leftrightarrow\) (iii) \(\Leftrightarrow\) (iv)

The given inequality is x + y \(\le\) 6
Draw the graph of the line x + y = 6.

Table of value satisfying the equation x + y = 6

Putting (0, 0) in the given inequation, we have
0 + 0 \(\le\) 6 \(\Rightarrow\) 0 \(\le\) 6, which is true
Half plane of x +y \(\le\) 6 is towards origin
Also the given unequality is x +y \(\ge\)4
Draw the graph of the line x + y = 4
Table of values satisfying the equation x + y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 \(\ge\) 4 \(\Rightarrow\) 0 \(\ge\) 4, which is false
Hald plane of x +y \(\ge\) 4 is awy from origin

(i) Here f(x) = - IxI
The function is defined for all real values of x. 
\(\therefore\) Domain of function = R
Now when x < 0, Ix I = - x
\(\therefore\) f(x) = - (- x) = x < 0
When x = 0, IxI = 0
\(\therefore\) f(x) = - I0I = 0​​​​​​​
When x> 0, IxI = x​​​​​​​
\(\therefore\) f(x) = - x < 0
So f(x) \(\le\)0 for all real values of x.
\(\therefore\) Range of function = (-\(\infty\), 0).
​​​​​​​(ii) f(x) = \(\sqrt{9-x^{2}}\) 
Since \(\sqrt{9-x^{2}}\) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].
For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.
∴ The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].